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Area of triangle ADB..... : Quantitative Comparison Questions

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Re: Area of triangle ADB..... [#permalink]
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Solving for x we see than ABC is a right triangle with 90 degree at A, 30 degrees at B and 60 at C.

Let side AC = 1 then side AB = \(\sqrt{3}\) and side BC= 2.

DC = 1 since ADC is an equilateral triangle




Area of ADC = \(\sqrt{3}\frac{1}{4}* (1)^2\)=\(\frac{1}{4}*\sqrt{3}\).


Area of big triangle ABC= \(\frac{1}{2}*1*\sqrt{3}\).

Area of triangle ABD = Area of big triangle ABC - Area of ADC = \(\frac{1}{2}*1*\sqrt{3}-\)\(\frac{1}{4}*\sqrt{3}\)=\(\frac{1}{4}*\sqrt{3}\).

Hence option C is the right answer.

PS: phoenixio's method is much better.
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Re: Area of triangle ADB..... [#permalink]
How is it possible for sides CD = BD when angle x corresponds to CD and angle 2x corresponds to BD? Kudos for explanation.
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Re: Area of triangle ADB..... [#permalink]
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In triangle \(ADC\), when solved for \(x\), the triangle will be an equilateral one.

Hence, \(AD = AC = DC\)

Now, in triangle \(ABD\)

\(AD\) & \(BD\) are the sides opposite to the same angle \(x\), that's why they will be same.

So, \(AD = BD\)

Finally, \(AD = BD = AC = DC\)

rubytuesdays21 wrote:
How is it possible for sides CD = BD when angle x corresponds to CD and angle 2x corresponds to BD? Kudos for explanation.
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Re: Area of triangle ADB..... [#permalink]
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Re: Area of triangle ADB..... [#permalink]
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